I'm reading "Journey Through Genius, the great theorems of mathematics" by William Dunham. During the introduction of Fermat's little theorem, it explains how Euler first proved it in 1736. However I couldn't understand one step of it.

3 Answers
3

One way is to look at the expression $\displaystyle \frac{p(p-1)(p-2)\cdots(p-n+1)}{1.2.\cdots.n}$ combinatorially as $\displaystyle \binom pn$. This represents the number of ways that you can choose $n$ objects out of $p$ objects when the order does not matter. Therefore it must be a natural number. Now, just by expanding $\displaystyle \binom pn$ for $n=1,\cdots,p-1$ you can factor out $p$ and obtain your expression by using the binomial theorem as $\displaystyle (a+1)^p - a^p$

EDIT: I guess you don't have a clear idea of what is happening here. Binomial coefficients, denoted by $\displaystyle \binom kn$ are always positive integers for the reasons that I said. Primality of $p$ comes into the play when you want to factor $p$ out of $\displaystyle \binom pn$. Since $n<p$ and $p$ is prime, $n!$ does not contain $p$ in its product. Therefore, the $p$ in the numerator does NOT get canceled and you can factor it out. That's why Fermat's little theorem is not correct for other natural numbers (A generalization exists due to Euler which uses Euler's totient function).
This proof is actually using mathematical induction on $a$. You prove that if its true for $a$, then it must be true for $a+1$ as well.

$\begingroup$please notice $p$ is a prime is a must.$\endgroup$
– athosSep 14 '13 at 1:59

$\begingroup$@athos: Nope. $p$ being a prime is not necessary for $\binom kn$ to be an integer. $p$ being prime is however a must for Fermat's little theorem to be true.$\endgroup$
– user66733Sep 14 '13 at 2:05

If $r$ is a real number such that $ra$ and $rb$ are both integers for
integers $a,b$ which are relatively prime (i.e. $(a,b) = 1$), then $r$ is an integer.

Now $\displaystyle r = \frac{(p-1)(p-2)\dots (p-k)}{(k+1)!}$ has the property that $pr$ and $(p-k-1)r$ are both binomial coefficients, and so are integers. Thus $r$ is an integer, by the above theorem.

To prove the above theorem, since $(a,b) = 1$, we have $ax + by = 1$ for some integers $x,y$. Multiplying by $r$ gives the result.